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Use the limit definition to find the equation for the slope of any tangent line, the slope of the tangent line at the given value, and the equation of the tangent line at the given value of x.

Use the limit definition to find the equation for the slope of any tangent line, the-example-1
User Ramakrishna Joshi
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2 Answers

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Final answer:

To calculate the slope of a tangent line at a specific point, use the slope formula with the provided points on the line, then use the point-slope form to find the equation of the tangent line using the calculated slope and a point on the line.

Step-by-step explanation:

To find the slope of a tangent line to a curve at a specific point, you can use the limit definition from calculus. This involves selecting two points on the tangent line, which is the same as the slope of the curve at that point, and using the formula:

  • slope (v) = (Y2 - Y1) / (X2 - X1)

where (X1, Y1) and (X2, Y2) are the coordinates of the two points on the tangent line. In the provided information, the endpoints of the tangent line correspond to positions of 1300 m at 19 s and 3120 m at 32 s. Plugging these into the formula gives the slope at t = 25 s.

To get the equation of the tangent line, you then use the point-slope form of a line:

y - y1 = m(x - x1)

where m is the slope and (x1, y1) is the point through which the line passes. Using the slope calculated earlier and the coordinates from the given point (t = 25 s), you can construct the equation of the tangent line.

User Arnaud Quillaud
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Answer:

Given function is:


f(x)=(3)/(x)\text{ at x=3}

To find the slope of any tanget line using limit definition.

we get,

Slope m is


m=\lim _(h\to0)(f(3+h)-f(3))/(h)
=\lim _(h\to0)((3)/(3+h)-1)/(h)
=\lim _(h\to0)(3-3-h)/(h(3+h))
=\lim _(h\to0)(-h)/(h(3+h))
=\lim _(h\to0)(-1)/(3+h)
=-(1)/(3)

Slope is -1/3

f(x) at x=3 is,

Put x=3 in f(x), we get


\begin{gathered} f(3)=(3)/(3)=1 \\ f(3)=1 \end{gathered}

we know that, the equation of line with slope m and point (x1,y1) is given is


y-y1=m(x-x1)

Here m=-1/3 and (x1,y1)=(3,1)

Substitute we get,


y-1=-(1)/(3)(x-3)
y-1=-(1)/(3)x+1
y=-(1)/(3)x+2

The required equation of a tangent line at x=3 is y=-1/3 x+2.

User Manu K Mohan
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